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Answer: The length and width of the rectangle are 19 cm and 5 cm respectively.
Step-by-step explanation: Given hat the length of a rectangle is four centimeters more than three times its width and the area of the rectangle is 95 square centimeters.
We are to find the length and width of the rectangle.
Let W and L denote the width and the length respectively of the given rectangle.
Then, according to the given information, we have
[tex]L=3W+4~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]
Since the area of a rectangle is the product of its length and width, so we must have
[tex]A=L\times W\\\\\Rightarrow 95=(3W+4)W\\\\\Rightarrow 3W^2+4W-95=0\\\\\Rightarrow 3W^2+19W-15W-95=0\\\\\Rightarrow W(3W+19)-5(3W+19)=0\\\\\Rightarrow (W-5)(3W+19)=0\\\\\Rightarrow W-5=0,~~~~~3W+19=0\\\\\Rightarrow W=5,~-\dfrac{19}{3}.[/tex]
Since the width of the rectangle cannot be negative, so we get
[tex]W=5~\textup{cm}.[/tex]
From equation (i), we get
[tex]L=3\times5+4=15+4=19~\textup{cm}.[/tex]
Thus, the length and width of the rectangle are 19 cm and 5 cm respectively.
The length of the rectangle is 19 and the width is 5 and it can be determined by using the formula of area of the rectangle.
Given that,
The length of a certain rectangle is four centimeters more than three times its width.
If the area of the rectangle is 95 square centimeters,
We have to determine,
The length and width of the rectangle.
According to the question,
Let the length of the rectangle be L,
And the width of the rectangle is W.
The length of a certain rectangle is four centimeters more than three times its width.
The perimeter of a square is the sum of the length of all its four sides.
The perimeter formulas of different two-dimensional shapes:
Then,
[tex]\rm L = 3W+4[/tex]
And If the area of the rectangle is 95 square centimeters,
The area of any polygon is the amount of space it occupies or encloses.
It is the number of square units inside the polygon.
The area is a two-dimensional property, which means it contains both length and width
[tex]\rm Area \ of \ the \ rectangle = length \times width\\\\L\times W = 95[/tex]
Substitute the value of L from equation 1,
[tex]\rm L\times W = 95\\\\(3W+4) \times W = 95\\\\3W^2+4W=95\\\\3W^2+4W-95=0\\\\3W^2+19W-15W-95=0\\\\W(3W+19) -5(3W+19) =0\\\\(3W+19) (W-5) =0\\\\W-5=0, \ W=5\\\\3W+19=0, \ W = \dfrac{-19}{3}[/tex]
The width of the rectangle can not be negative than W = 5.
Therefore,
The length of the rectangle is,
[tex]\rm L = 3W+4\\\\L = 3(5)+4\\\\L=15+4\\\\L=19[/tex]
Hence, The length of the rectangle is 19 and the width is 5.
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